How do you solve #root4(2x)+root4(x+3)=0#?

1 Answer
Apr 29, 2017

Answer:

This equation has no solutions.

Explanation:

Note that #root(4)(t) >= 0# for all #t >= 0# and has no real value when #t < 0#.

So the only circumstances under which the left hand side of the given equation can be #0# are if both:

#2x=0" "# and #" "x+3=0#

The first of these implies that #x=0# and the second that #x=-3#.

These are incompatible, so there are no real solutions.

#color(white)()#
Complex solutions?

Given:

#root(4)(2x) + root(4)(x+3) = 0#

Subtract #root(4)(x+3)# from both sides to get:

#root(4)(2x) = -root(4)(x+3)#

Raise both sides to the #4#th power (noting that this is where we might introduce extraneous solutions) to get:

#2x = x+3#

Subtract #3# from both sides to find:

#x = 3#

But...

#root(4)(2(color(blue)(3)))+root(4)(color(blue)(3)+3) = root(4)(6)+root(4)(6) != 0#

So there are no solutions at all.